Question

find t and the terminal point determined by t for each point in the following figure, where t is increasing in increments of π/6. (order your answers from smallest to largest t.)

Explanation:

Step1: Recall the unit - circle terminal - point formula

For a unit circle \(x = \cos t\) and \(y=\sin t\). When \(t = 0\), \(\cos(0)=1\) and \(\sin(0) = 0\).

Step2: Analyze the increment of \(t\)

Since \(t\) is increasing in increments of \(\frac{\pi}{6}\).
When \(t=\frac{\pi}{6}\), \(\cos(\frac{\pi}{6})=\frac{\sqrt{3}}{2}\) and \(\sin(\frac{\pi}{6})=\frac{1}{2}\), the terminal point is \((\frac{\sqrt{3}}{2},\frac{1}{2})\).
When \(t = \frac{\pi}{3}\), \(\cos(\frac{\pi}{3})=\frac{1}{2}\) and \(\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\), the terminal point is \((\frac{1}{2},\frac{\sqrt{3}}{2})\).
When \(t=\frac{\pi}{2}\), \(\cos(\frac{\pi}{2}) = 0\) and \(\sin(\frac{\pi}{2})=1\), the terminal point is \((0,1)\).
When \(t=\frac{2\pi}{3}\), \(\cos(\frac{2\pi}{3})=-\frac{1}{2}\) and \(\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\), the terminal point is \((-\frac{1}{2},\frac{\sqrt{3}}{2})\).
When \(t=\frac{5\pi}{6}\), \(\cos(\frac{5\pi}{6})=-\frac{\sqrt{3}}{2}\) and \(\sin(\frac{5\pi}{6})=\frac{1}{2}\), the terminal point is \((-\frac{\sqrt{3}}{2},\frac{1}{2})\).
When \(t=\pi\), \(\cos(\pi)=- 1\) and \(\sin(\pi)=0\), the terminal point is \((-1,0)\).
When \(t=\frac{7\pi}{6}\), \(\cos(\frac{7\pi}{6})=-\frac{\sqrt{3}}{2}\) and \(\sin(\frac{7\pi}{6})=-\frac{1}{2}\), the terminal point is \((-\frac{\sqrt{3}}{2},-\frac{1}{2})\).
When \(t=\frac{4\pi}{3}\), \(\cos(\frac{4\pi}{3})=-\frac{1}{2}\) and \(\sin(\frac{4\pi}{3})=-\frac{\sqrt{3}}{2}\), the terminal point is \((-\frac{1}{2},-\frac{\sqrt{3}}{2})\).
When \(t=\frac{3\pi}{2}\), \(\cos(\frac{3\pi}{2}) = 0\) and \(\sin(\frac{3\pi}{2})=-1\), the terminal point is \((0, - 1)\).
When \(t=\frac{5\pi}{3}\), \(\cos(\frac{5\pi}{3})=\frac{1}{2}\) and \(\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}\), the terminal point is \((\frac{1}{2},-\frac{\sqrt{3}}{2})\).
When \(t=\frac{11\pi}{6}\), \(\cos(\frac{11\pi}{6})=\frac{\sqrt{3}}{2}\) and \(\sin(\frac{11\pi}{6})=-\frac{1}{2}\), the terminal point is \((\frac{\sqrt{3}}{2},-\frac{1}{2})\).

Answer:

\(t\)	Terminal Point
\(\frac{\pi}{6}\)	\((\frac{\sqrt{3}}{2},\frac{1}{2})\)
\(\frac{\pi}{3}\)	\((\frac{1}{2},\frac{\sqrt{3}}{2})\)
\(\frac{\pi}{2}\)	\((0,1)\)
\(\frac{2\pi}{3}\)	\((-\frac{1}{2},\frac{\sqrt{3}}{2})\)
\(\frac{5\pi}{6}\)	\((-\frac{\sqrt{3}}{2},\frac{1}{2})\)
\(\pi\)	\((-1,0)\)
\(\frac{7\pi}{6}\)	\((-\frac{\sqrt{3}}{2},-\frac{1}{2})\)
\(\frac{4\pi}{3}\)	\((-\frac{1}{2},-\frac{\sqrt{3}}{2})\)
\(\frac{3\pi}{2}\)	\((0,-1)\)
\(\frac{5\pi}{3}\)	\((\frac{1}{2},-\frac{\sqrt{3}}{2})\)
\(\frac{11\pi}{6}\)	\((\frac{\sqrt{3}}{2},-\frac{1}{2})\)